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Post by karfston on Dec 11, 2019 17:44:45 GMT
Let’s say I have an open-ended cylinder. It’s walls emit... gama? X-ray? UV? Microwave? Perhaps it’s that green stuff that signals ‘poison/radiation/biohazard’ in movies. Regardless, it’s some form of radiation obeying the inverse-square law that I wish to avoid.
Where do I need to be to minimize that dose while passing through the cylinder? (I’m extrapolating slightly from other knowledge I’ve gleaned over the years, but I’m guessing being in the center doesn’t help... heck, being in the center may actually make things worse if the radiation is being emitted from inside the walls as opposed to on their edge).
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Post by pencilcrayonphoenix on Dec 11, 2019 21:07:45 GMT
My guess is "be in the center" is less "you won't get a lethal dose of radiation" and more "we can keep you from dying for long enough to be useful, and we might not even need to haul out your backup afterwards".
They did list this as "likely a suicide mission", after all. It's just that death's becoming more of a video game style 'time to respawn' type death for society, with the Laz5 option, rather than "well, they're gone Forever".
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Post by karfston on Dec 12, 2019 7:02:31 GMT
So, here’s the thing. The further I get from one wall, the closer I get to all the other wall. My general understanding is, if I’m inside a hollow sphere, I’m unaffected by the gravitational pull of thesphere’s walls. en.wikipedia.org/wiki/Shell_theoremSubstitution 1: light and gravity both obey the inverse square law - the same math applies. A uniformly lit sphere should have constant brightness across its interior. Substitution 2: replacing a sphere with a cylinder, which I must then traverse from one end to the other. I’m looking at a special case here, but I’d expect for this special case (ignoring the ends of the cylinder, treating it as one of infinite length, or alternately as one that I must traverse it’s length but not it’s radius...) that it doesn’t matter where I am relative to the walls - the lighting/radiation is of equal intensity. If so, I see no merit in avoiding the walls. |
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Post by freeflier on Dec 12, 2019 13:59:38 GMT
Special case: touching the walls may result in contamination, which would continue the exposure after passage was complete.
Also, in an open-ended cylinder, some of the radiation will escape out the ends.
The further from the walls you are, the more will escape, so I would expect the path of least exposure to be right through the center.
Gripping hand, radiation falls off with the square of the distance . . . double the distance is one-quarter the exposure, so again, right through the center.
And finally, through the center means less chance of contact with the walls, which allows higher speeds, which means less exposure.
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Post by karfston on Dec 13, 2019 3:49:48 GMT
I’ve been thinking on this... but also, I think the math is slipping past some people.
It’s been too long since I used calculus of sufficient difficulty to keep my skills up to date, so I can’t work the proof without serious refresher, but I believe the squared distance to all points along the wall is a constant for any point inside that circle - the same constant for all points. When I’m agains t the edge of a circle and back up, I’m moving away from a tiny piece of its edge, and moving towards a huge bit of it’s edge (more than half).
Taking the full of the scenario in the comic changes the equation some. A hole was just blasted in the wall. This corresponds to a huge transfer of energy. Most of the hull at the center will have been vaporized. More will have been rendered molten and sent away at high speed. Further from the center, material will have suffered more generic kinetic failures - failures induced not by heat but by the shockwave of the blast.
Working this backwards outside in, we have fractured material coated with molten material (probably radioactive) and an ever-present vapor (definitely radioactive). Going through that hole seconds after it’s made, you’re going to come into contact with that vapor component, and at least some of it’s going to deposit on any surfaces that come near it.
‘Scritchy’ is correct. The sensors or their housings probably all picked up a coating of radioactive vapor-deposit and are reporting a significant noise-level from that deposit.
Direction to ‘keep clear of the radioactive walls’ may or may not directly help, but it will motivate people to not linger where said vapors can still deposit.
—
Sad as it is, my formal math studies went up to calc 3. My formal math needs at work include some trig, a bit of Calc 1, and tons of statistics, probability, and forecasting. Correct Binomial math is hard. Oh right, solving a 52x52 matrix? Not something your beginner’s guide to matrices cover.
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Post by sotanaht on Dec 13, 2019 11:58:02 GMT
Just logic check it. If the hole were arbitrarily large (radius), would any radiation reach the center? No, at some point the distance from the nearest wall is far enough that decay (if nothing else) won't allow the radiation to travel that far. Of course with a small enough diameter there's another case to consider. Imagine you were to make a ring of about a dozen or so lasers all pointing towards a single spot in the center, where would the light be brightest, at one of the lasers, or at the intersection of all of them? If the radiation is being projected from the walls, it's possible that the intersection point (center) would be worst, but only if the distance from the center to the walls is small enough that most of the radiation isn't lost in transit.
Conclusion: It depends on the properties of the radiation and the radius of the circle, but I doubt there is any possible scenario where the radiation is exactly equal at all points inside the circle.
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Post by karfston on Dec 16, 2019 7:06:19 GMT
Just logic check it. If the hole were arbitrarily large (radius), would any radiation reach the center? No, at some point the distance from the nearest wall is far enough that decay (if nothing else) won't allow the radiation to travel that far. Checking logic. Radiation takes 2 general forms. Electromagnetic. Range: Only limited by the Hubble radius (billions of light years). Particle radiation: Alpha particles: Range - unlimited. Common name: Helium. Beta particles: Range - notionally unlimited - half-life: not less than 6.6 x 10^28 years. Common name: Electron/Positron (depending on flavor) Gamma: See above, electromagnetic. Common name: Photon. Neutron: Range: Short (when compared to other types). Half-life: 10.3 minutes. Anti-protons: Given the weapon in use, there might be some of this stuff lingering. Half-life: Theoretically stable, experimentally at least 1 million years. Unless we’re slinging neutrons around slow enough for their half-life to matter, or using distances where Hubble has some input, that radiation will make it to the center. —- Consider, if you would, an empty box. Let’s make the sides on this box 1 foot each, and have them emit a steady and uniform light of 1 lumen. In the middle of the box, we’ll place a light sensor. No matter where I move the sensor in the box, it’s always going to be exposed to the same light level. It’s no brighter in the corners than it is in the middle, or along any side. No matter how much larger I make the box (10 feet, 100 meters, 100 lightyears), that light sensor will still pick up the same amount of light, no matter where that sensor is in the box. These are simple properties of light - the brightness is reduced by 1/d-squared, but the light emitting surface area per steradian is d-squared. Summary：d-squared over d-squared is 1, distance has no effect. |
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